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Hyperbolic Geometry Lecture 2 John Stogin October 3, 2009 First, we begin with a couple comments from last week. There are two matrices in SL(2, R) that correspond with the same isometry of H, and their coordinates diﬀer by a sign. So we will work with P SL(2, R) = SL(2, R)/±12 as our set of isometries of H. Also, P SL(2, R) is a subset of a special class of functions called conformal maps. These maps satisfy the Cauchy-Riemann ∂v ∂v equations given by ∂u = ∂y and ∂u = − ∂x for u + iv = T (x + iy). They also ∂x ∂y have the property that their derivative is never zero and they preserve angles. 1 Hyperbolic Geometry on the Unit Disk 1.1 Introducing U Last week, we saw how the hyperbolic geometry can be represented as a half 2 +dy 2 plane, H = {z ∈ C|Im(z) > 0} with a metric ds2 = dx y2 . Today, we will also consider another representation. Deﬁnition (The Unit Disk) We call U = {z ∈ C||z| < 1} the unit disk. There is a conformal map f : H → U given by f (z) = zi+1 . Note that this is a z+i bijection and maps the boundary R ∪ {∞} of H to the boundary {z||z| = 1} of U. We also deﬁne a distance du on U by du (z0 , z1 ) = dh (f −1 z0 , f −1 z1 ) where dh is the distance on H. 4|dz|2 Theorem 1.1 The distance du can be calculated with the metric ds2 = (1−|z|2 )2 Proof We will show this by calculating the length of a curve γ : [a, b] → U. −1 b | d f ◦γ(t)|dt So lu (γ) = lh (f −1 ◦ γ) = a dt Im(γ) . In order to do this, we must −1 ﬁrst calculate f and we do so by solving w = f (z) = zi+1 for z. So z+i wz + iw = zi + 1 ⇒ z(w − i) = −iw + 1 ⇒ z = f −1 (w) = −iw+1 . Now, we cal- w−i −iγ+1 culate the denominator of the integral expression. Im(f −1 γ) = Im γ−i = (−iγ+1)(¯ +i) γ Im(−iγ γ +γ+¯ +i) ¯ γ 1−|γ|2 Im = |γ−i|2 = |γ−i|2 . And the numerator becomes (γ+i)(γ+i) d −1 −1 dt f ◦ γ(t) dt = |(f ) ◦ γ(t)||γ (t)|dt and for the sake of simplicity, we will now evaluate d −1 dz f (z) = d dz −iz=1 z−i = −i(z−i)−(−iz+1) = (z−i)2 . So the entire (z−i)2 −2 b 2 b 2|dz|2 |γ−i| integral expression now becomes a 2|γ (t)|dt 1−|γ|2 = a 1−|z|2 where z = γ(t). |γ−i|2 For an example, let us calculate verify the following corollary. 1 Corollary 1.2 The distance between a point in U with radius r and the origin r+1 is ln r−1 . Proof We can choose our point to be a real positive number without any r 2dx r 1 1 r loss of generality. du (0, r) = 0 1−x2 = 0 1−x − 1+x dx = [− ln |1 − x|]0 + r 1+r [− ln |1 + x|]0 = ln 1−r . 1.2 Two related lemmas |z−w|2 Lemma 1.3 For z, w ∈ H, cosh(dh (z, w)) = 1 + 2Im(z)Im(w) . b Proof We know dh (ai, bi) = ln a assuming without loss of generality that a < b. 2 +a2 Let’s verify the formula in this case. cosh(dh (ai, bi)) = b/a+a/b = b 2ab = 2 2 2 2 |ai−bi| 2ab 2ab + b −2ab+a = 1 + 2Im(ai)Im(bi) . Now, we know that distance is invari- 2ab ant under Mobius transformations, so if the right hand side is also invariant, then we can prove the equation for any z and w by selecting a Mobius trans- formation T mapping both of them to the imaginary axis. First let’s evaluate 2 2 az+b aw+b adz+bcw−bcz−adw |z−w|2 |T z − T w|2 = cz+d − cw+d = (cz+d)(cw+d) = |cz+d|2 |cw+d|2 . At this point, we only have to evaluate Im(T z) = Im az+b cz+d z +d) = Im (az+b)(c¯+d) = z (cz+d)(c¯ ¯ z Im(abz z +adz+ad¯+bd) |cz+d|2 z = Im(adz+bc¯) = |cz+d|2 (ad−bc)Im(z) |cz+d|2 Im(z) = |cz+d|2 . We now see that we will get proper cancellation and that the right side of our equation is also invariant. 2|z−w|2 Lemma 1.4 In U, lemma (1.3) becomes cosh(du (z, w)) = 1 + (1−|z|2 )(1−|w|2 ) . Proof The proof of this lemma is similar to the proof of lemma (1.3). It is not diﬃcult to show that f −1 (z) = −iz+1 . From lemma (1.3), cosh(du (z, w)) = z−i |f −1 z−f −1 w|2 cosh(dh (f −1 z, f −1 w)) = 1 + 2Im(f −1 z)Im(f −1 w) . We can evaluate the numera- 2 2 −iz+1 −iw+1 tor of this expression: |f −1 z − f −1 w|2 = z−i − w−i = 2w−2z (z−i)(w−i) = 4|z−w|2 At this point, we only have to evaluate Im(f −1 z) to ensure that |z−i|2 |w−i|2 . we will get proper cancellation in our fraction. Let’s do that. Im(f −1 z) = −iz+1 z (−iz+1)(¯+i) ¯ z Im(−iz z +z+¯+i) 1−|z|2 Im z−i = Im = |z−i|2 = |z−i|2 , which is what we (z−1)(z−i) want. 2 Hyperbolic Trigonometry 2.1 Angle of Parallelism Consider a triangle ∆ in H with a vertex at ∞ and another vertex and right angle at i. Deﬁnition We deﬁne Π(a) to be the third angle of ∆ with a side of length a opposite the vertex at ∞. We can now explore the following relations. 2 Theorem 2.1 Given ∆ satisfying the conditions mentioned above, the following three relations hold. 1 (i) tan Π(a) = sinh(a) 1 (ii) sin Π(a) = cosh(a) 1 (iii) sec Π(a) = tanh(a) Proof From lemma (1.3), let the third vertex of ∆ be v so we can validate (ii). Π(a) |i−v|2 4 sin2 ( π − 2 ) 4 sin Π(a)+(1−cos( π −Π(a))) 2 cosh(a) = 1 + 2Im(i)Im(v) = 1 + 2 sin Π(a) = sin Π(a) = 1 sin Π(a) . 1 To validate (iii), note sec Π(a) = cos Π(a) = 1 − sin2 Π(a) = 1− 1 cosh2 (a) = √ cosh2 (a)−1 sinh(a) cosh(a) = cosh(a) = tanh(a). Now, (i) is trivial. 2.2 The Sine and Cosine Rules In our new hyperbolic trigonometry, there are some formulae that are similar to those that we have seen in trigonometry. As usual in trigonometry, we concern ourselves with the study of triangles with sides measuring a, b, and c, and having opposite angles α, β, and γ respectively. Theorem 2.2 (Cosine Rule I) cosh(c) = cosh(a) cosh(b)−sinh(a) sinh(b) cos(γ) Proof For this proof, we will deﬁne a triangle ∆ in U with vertices vc = 0, r = Re(va ) and Im(va ) = 0, and z = vb and opposite lengths c, a, and b respec- tively. Note that there is no loss of generality here. From lemma (1.4), cosh(c) = 2|va −vb |2 (1−|va |2 )(1−|vb |2 ) + 1. We will now verify the right side matches this expression. 1+r 1−r 1−r + 1+r 2(1+r 2 ) Note using corollary (1.2), cosh(b) = cosh(du (0, r)) = 2 = 2(1−r 2 ) = 2 1+r 2 2 1−r 2 . Now we can also ﬁnd sinh(b) = cosh2 (b) − 1 = 1+r 1−r 2 −1 = √ (1+r 2 )2 −(1−r 2 )2 2r 1−r 2 = 1−r2 . And from the traditional cosine rule, |z − r|2 = |z|2 + 2 r −2|z|r cos(γ). Now, we can directly calculate the right side. cosh(a) cosh(b)− 2 2 2 2 1+r 2 1+|z| sinh(a) sinh(b) cos(γ) = 1−r2 1−|z|2 − (1−r2 )(1−|z|2 ) r +|z| −|z−r| 4r|z| 2r|z| 2 2 2 2 2 r 2 |z|2 −r 2 −|z|2 +1+2|z−r|2 2|z−r|2 = (1+r )(1+|z| )−2r −2|z| +2|z−r| = (1−r 2 )(1−|z|2 ) (1−r 2 )(1−|z|2 ) = 1+ (1−r 2 )(1−|z|2 ) , which is what we wanted. sinh(a) sinh(b) Theorem 2.3 (Sine Rule) sin(α) = sin(β) 2 sinh(c) sinh2 (c) sinh2 (c) Proof sin(γ) = 1−cos2 (γ) = 2 from theorem (2.2). cosh(a) cosh(b)−cosh(c) 1−( ) sinh(a) sinh(b) We will now show the right side is symmetric in a, b, and c. It can be rewrit- sinh2 (a) sinh2 (b) sinh2 (c) ten as sinh2 (a) sinh2 (b)−(cosh(a) cosh(b)−cosh(c))2 , which is symmetric provided the denominator is symmetric. Using cosh2 (x) = 1 + sinh2 (x) and expanding, we 3 see the denominator is equivalent to sinh2 (a) sinh2 (b) − cosh2 (a) cosh2 (b) − cosh2 (c) + 2 cosh(a) cosh(b) cosh(c) = sinh2 (a) sinh2 (b)−(1+sinh2 (a)) cosh2 (b)−cosh2 (c)+2 cosh(a) cosh(b) cosh(c) = sinh2 (a)(sinh2 (b) − cosh2 (b)) − cosh2 (b) − cosh2 (c) + 2 cosh(a) cosh(b) cosh(c). Since sinh2 (b) − cosh2 (b) = −1, we are done. cos(α) cos(β)+cos(γ) Theorem 2.4 (Cosine Rule II) cosh(c) = sin(α) sin(β) Proof This proof is rather straightforward and tedious. It is therefore left as an exercise. Note that this third theorem has no analogue in trigonometry on the Euclidean plane. Can you see why? It determines the length of a side from the three angles, whereas in Euclidean trigonometry, there are many diﬀerent triangles with congruent angles. 3 Hyperbolic Area Since we have explored the notion of length, and we are working in a two dimensional space, it seems only natural to explore the concept of area next. dxdy Deﬁnition (Area) For a region A ⊂ H, deﬁne the area to be µ(A) = A y2 if the integral exists. This is a natural deﬁnition of area as we will see in the following theorem. Theorem 3.1 The area µ(A) is invariant under transformations in P SL(2, R). I.e. for T ∈ P SL(2, R), µ(T (A)) = µ(A). az+b Proof First deﬁne T (z) = cz+d where ad−bc = 1. Let w = u+iv = T (z). Then ∂u ∂v ∂u ∂v ∂u 2 ∂v 2 ∂ 2 dudv = ∂x ∂y − ∂y ∂x dxdy = ∂x + ∂x dxdy = ∂x (u + iv) dxdy = 2 2 ∂ ∂x T z(x) dxdy = dT dxdy = |cz + d|−4 dxdy. The second equality holds by dz the Cauchy-Reimann equations. Then, recalling our proof of lemma (1.3) that Im(z) v = Im(T (z)) = |cz+d|2 , we calculate dudv |cz+d|−4 dxdy dxdy µ(T (A)) = T (A) v2 = A Im(z)2 |cz+d|−4 = A y2 = µ(A) 4 Gauss-Bonnet Theorem Theorem 4.1 (Gauss-Bonnet) Let ∆ be a triangle with angles α, β, and γ. Then µ(∆) = π − α − β − γ. Proof First, we will observe the case where a vertex of the triangle lies on the boundary ∂H and from theorem (3.1) use a mobius transformation to map this vertex to ∞. This yields a region bounded on the left and right by two vertical lines and below by a subset of a circle. Without changing angles and area, we may apply transformations of the form T (z) = z + b or T (z) = az so that this circle is the unit circle. (Let the left side be adjacent to angle α and have horizontal coordinate a, while the right side is adjacent to angle β and has horizontal coordinate b.) Finally, we have a region over which it is not diﬃcult 4 to integrate. b ∞ b µ(A) = A dxdy = y2 a dx √ dy 1−x2 y 2 = a √ dx 1−x2 With a change of coordinates β − sin(θ) (x = cos(θ)), the last integral becomes π − α − β. π−α sin(θ) = Note that in this case, γ = 0. We are now left with the case where no vertex lies on ∂H. To handle this, we simply reduce this to two instances of our previous case by extending one of the line segments to a ray. We form two new triangles and take their diﬀerence to be the area of the original triangle. 5 Various Trigonometries We will end this lecture with a few comments on trigonometries in various diﬀerent geometries. Those are the spherical geometry, the hyperbolic geometry and the Euclidean geometry. In all three cases, we deal with triangles with sides measuring a, b, and c, and having opposite angles α, β, and γ respectively. 5.1 Triangles on the Sphere For a triangle on a sphere with radius r, the three trigonometric identities take the following form. sin(a/r) sin(b/r) Sine Rule: = (1) sin(α) sin(β) Cosine Rule I: cos(c/r) = cos(a/r) cos(b/r) + sin(a/r) sin(b/r) cos(γ) (2) cos(α) cos(β) + cos(γ) Cosine Rule II: cos(c/r) = (3) sin(α) sin(β) 5.2 Triangles on the Hyperbolic Plane To convert the previous equations into Hyperbolic form, we use the following deﬁnition. Informally, it may be thought of as the “sphere with imaginary radius.” r 2 (dx2 +dy 2 ) Deﬁnition Let Hr be the upper half plane with metric ds2 = y2 . Note that letting r = 1, we get the usual hyperbolic plane. We now take the three previous equations and replace r with ir. Using the identities sin(ix) = i sinh(x) and cos(ix) = cosh(x), we obtain the following hyperbolic trigonomet- ric identities. sin(a/r) sin(b/r) Sine Rule: = (4) sin(α) sin(β) Cosine Rule I: cosh(c/r) = cosh(a/r) cosh(b/r) + sinh(a/r) sinh(b/r) cos(γ) (5) cos(α) cos(β) + cos(γ) Cosine Rule II: cosh(c/r) = (6) sin(α) sin(β) 5 5.3 Trigonometry on the Euclidean Plane We can convert the above equations to Euclidean trigonometric equations by taking the second order Taylor expansion of sin, cos, sinh, and cosh and letting a, b, c, r → ∞. Equation (4) becomes a b = (7) sin(α) sin(β) which is the standard Sine Rule. Likewise, equation (5) becomes c2 = a2 + b2 − 2ab cos(γ) (8) which is the standard Cosine Rule. As mentioned earlier, there is no analogue to the Cosine Rule II. 6